Divide x^3 - 4x + 5 by x^2 + 2. Give the quotient and the remainder.

Step-by-step solution

1. 1

   Write both polynomials with every power shown. The dividend has no $x^2$ term and the divisor has no $x$ term:

   $\text{dividend}=x^3+0x^2-4x+5,\qquad \text{divisor}=x^2+0x+2$

   Because the divisor has degree $2$, synthetic division does not apply and long division is the correct tool.

2. 2

   Get the first quotient term from the leading coefficients.

   $\frac{x^3}{x^2}=x$

   Multiply the divisor by $x$: $x\left(x^2+2\right)=x^3+2x$. Subtract it from the dividend:

   $\left(x^3+0x^2-4x+5\right)-\left(x^3+0x^2+2x\right)=-6x+5$

3. 3

   Check whether the algorithm can continue. The leftover $-6x+5$ has degree $1$, and the divisor has degree $2$. Since $1<2$ there is no further term to extract, so the process stops immediately after one step:

   $\text{quotient}=x,\qquad \text{remainder}=-6x+5$

   A single-term quotient is expected: the degrees differ by exactly one, so the quotient has degree $3-2=1$.

4. 4

   Write the two standard forms of the answer.

   $\frac{x^3-4x+5}{x^2+2}=x+\frac{-6x+5}{x^2+2}$

   $x^3-4x+5=\left(x^2+2\right)x+(-6x+5)$

   The second version is the identity to verify since it is fraction-free.

5. 5

   Verify, and note what the remainder tells you. Expanding, $\left(x^2+2\right)x=x^3+2x$, and $x^3+2x-6x+5=x^3-4x+5\;\checkmark$. Because the remainder is not zero, $x^2+2$ is not a factor — consistent with the fact that the roots of $x^2+2$ are $x=\pm i\sqrt2$, and substituting $x^2=-2$ into the dividend gives $x\cdot(-2)-4x+5=-6x+5\ne 0$, exactly the remainder.